Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading a proof from a paper I found online, and it goes like this: We want to show that there exists a surjection $f$ from the cantor set $\mathfrak{C}$ to the interval $[0,1]$ then we can show that $|\mathfrak{C}|\geqslant [0,1]$, but since $\mathfrak{C}\subseteq [0,1]$ it follows that $|[0,1]|=|\mathfrak{C}|$. Since the unit interval is uncountable, the Cantor set must be uncountable as well.

I am stuck on one word here:

After showing that for each $p\in \mathfrak{C}$ none of its digits can equal 1, we then define a function $f : C \to [0, 1]$ by taking the number not consisting entirely of the digits $\{0, 2\}$ (why not entirely?) and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation. To show that $f$ is surjective, consider any element in $a\in[0, 1]$. Represent $a$ in its binary form, and then replace each occurrence of $\{1\}$ digit by a $\{2\}$. This new number, which we label $b$, satisies $f(b) = a$. As a result, $f$ is surjective. And then it follows that follows that $|[0,1]|=|\mathfrak{C}|$.

I can't seem to see the necessity of the word entirely I can't see why the numbers $1=0.2222...$ and $0=0.000...$ would hurt anything, and $0,1\in\mathfrak{C}$, so if $f$ doesn't consider those numbers then it is an inconsistent mapping from $\mathfrak{C}$ to $[0,1]$. (by inconsistent I mean that $\text{Dom}f\neq \mathfrak{C}$).

share|cite|improve this question
I also fail to see why the author would want to say "entirely" there. Also it seems to break the proof, because the domain of the function $f$ would no longer be the entire Cantor set (as you point out in the last paragraph, although I wouldn't use the word "inconsistent.") – Trevor Wilson Oct 26 '13 at 2:25
@TrevorWilson There is a book Fondations of Abstract Analysis: Second Edition by Jewgeni H. Dshalalow, I got the term inconsistent function from there. He distinguishes be between a mapping $f:A\to B$, if $\text{Dom}f=A$ then $f$ is consistent, and if this is not the case then $f$ is not consistent. Later he discusses how we can make $f$ consistent by defining $f$ from the power set of $A$ to the power set of $B$. I haven't got to a part when this was useful yet (I only use it as a supplement to the book we use in our course). But, I like the authors precision, its worth checking out. – JimmyJackson Oct 26 '13 at 3:48 me the notation $f:A \to B$ means that $A$ is the domain of $f$. But maybe analysts do things differently. – Trevor Wilson Oct 26 '13 at 5:38
up vote 1 down vote accepted

I think what “taking the number not consisting entirely of the digits $\{0,2\}$ and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation” means it the following. Take any number $b\in\mathfrak C$. Then, the ternary representation of $b$ consists entirely of the digits $\{0,2\}$ (i.e., it does not contain any digit $1$). Let $a$ be the number constructed from $b$ in such a way that each digit $2$ in $b$ is replaced by $1$ in $a$ (and you leave the zeros alone). Define $f(b)=a$.

Then, take any $a\in[0,1]$. The aim to show that $a=f(b)$ for some $b\in\mathfrak C$. Consider the binary representation of $a$. Replace all digits $1$ in the binary representation in $a$ by $2$ and view this new number as a ternary representation. Call this new number $b$. Since $b$ consists entirely of the digits $\{0,2\}$, $b\in\mathfrak C$, and $f(b)=a$ by the construction of $f$. Hence, $f$ is surjective.

Upon a closer look, I might have misunderstood the source of the confusion (metaconfusion :-)). You may just need to replace “not consisting” by “consisting.” That is, you just consider those numbers whose ternary representations consist entirely of $\{0,2\}$ The set of such numbers is $\mathfrak C$, issues with uniqueness of representations aside.

share|cite|improve this answer
Thanks! That makes sense maybe the author accidentally placed the 'not' in front the consisting. Its amazing how one little word can have such a drastic effect on a proof, but I guess that what makes math so fun; nothing should be ambiguous. – JimmyJackson Oct 26 '13 at 3:34
@JimmyJackson I know, right? Even the most authoritative sources are full of typos (which is natural), but I usually get as confused as you did now when I read something that is obviously not true! I think we just sometimes need to rely on our common sense and not be afraid of mentally correcting the author when something looks really off. – triple_sec Oct 26 '13 at 3:38
I agree! But, I am not so confident in my mathematics. With time I will hopefully be able to make those corrections on my own: until then I am thankful to the math community here for all of its guidance. – JimmyJackson Oct 26 '13 at 3:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.